Question

for what value(s) of k will the function h(x) = 4x^2 -3x +2kx +1 have two zeros? I got +/- 4 but i think its wrong so just need some clarification please.

Answer 1

THIS IS EASSYYYY!!!!

Answer 2

h(x) has 2 zeroes when the discriminant is positive

Answer 3

0.o

Answer 4

h(x) = 4x^2 + (-3+2k)x + 1

Answer 5

I'm not stupid i know that too

Answer 6

D = b^2 - 4*a*c
D = (-3+2k) - 4*4*1 > 0
-3 + 2k - 16 > 0
-19 + 2k > 0

Answer 7

2k > 19
k > 19/2

Answer 8

Noo... its b^2 so (-3 + 2k) ^2 right?

Answer 9

awww shoot

Answer 10

:)

Answer 11

Thanks for catching the mistake

Answer 12

no probs

Answer 13

So then was i right?

Answer 14

D = (-3+2k)^2 - 4*4*1 = 9 -12k + 4k^2 - 16 = 4k^2 -12k - 7 >0

Answer 15

But then what does k have to be?

Answer 16

(2k-7)(2k+1)>0

Answer 17

wait but where'd you get 9-12k

Answer 18

(-3+2k)^2 - 4*4*1 =
(9 -12k + 4k^2) - 16 =
(2k-7)(2k+1)>0

Answer 19

Ohhh, lol thank you!

Answer 20

so then k has to be either 7/2 or -1/2 right?

Answer 21

well when is a product of two numbers positive?

Answer 22

when the numbers are negative?

Answer 23

or?

Answer 24

don't know

Answer 25

when both numbers have the same sign:
7*1 = 7
(-7)(-1) = 7

Answer 26

oh OR they're positive?

Answer 27

OMG YOU GOT IT!!!!

Answer 28

lol yay :D
but that is right though right?

Answer 29

Like how k has to be those two numbers?

Answer 30

Not yet.. so it means:
Case 1:
(2k-7) < 0, 2k < 7, k<7/2
(2k+1) < 0, 2k<(-1), k<(-1/2)

Case 2:
(2k-7) > 0, 2k > 7, k > 7/2
(2k+1) > 0, 2k > -1, k > (-1/2)